Begin2.DVI

Example 7-27. Consider a surface described by the implicit form F(x, y, z) = 0.

Recall that equations of this form define zas a function of xand yand the derivatives

of zwith respect to xand yare given by

∂z

∂x =

−Fx

Fz =

−∂F∂x

∂F

∂z

and

∂z

∂y =

−Fy

Fz =

−∂F∂y

∂F

∂z

Substituting these derivatives into the equation (7.86) and simplifying one finds that

the direction cosines of the unit normal to the surface are given by

nx=

∂F∂x

√

(∂F

∂x

) 2

+

(

∂F

∂y

) 2

+

(∂F

∂z

) 2

, n y=

∂F

√ ∂y

(∂F

∂x

) 2

+

(

∂F

∂y

) 2

+

(∂F

∂z

) 2

, n z=

∂F∂z

√

(∂F

∂x

) 2

+

(

∂F

∂y

) 2

+

(∂F

∂z

) 2

Tangent Plane to Surface

Consider a smooth surface defined by the equation

r =
r (u, v ) = x(u, v )ˆe 1 +y(u, v )ˆe 2 +z(u, v )ˆe 3

In order to construct a tangent plane to a regular point
r =
r (u 0 , v 0 )where the surface

coordinates have the values (u 0 , v 0 ), one must first construct the normal to the surface

at this point. One such normal is

N
 =∂
r

∂u ×

∂
r

∂v =N^1 ˆe^1 +N^2 ˆe^2 +N^3 ˆe^3

The point on the surface is

x 0 =x(u 0 , v 0 ), y 0 =y(u 0 , v 0 ), z 0 =z(u 0 , v 0 )

which can be described by the position vector
r 0 =
r (u 0 , v 0 ). If
r represents the

variable point

r =xˆe 1 +yˆe 2 +zˆe 3

which varies over the plane through the point (x 0 , y 0 , z 0 ), then the vector
r −
r 0 must

lie in the tangent plane and consequently is perpendicular to the normal vector N
.

One can then write

(
r −
r 0 )·N
 = 0 (7 .87)

as the equation of the plane through the point (x 0 , y 0 , z 0 )which is perpendicular to

N
and consequently tangent to the surface. In equation (7.87) one can substitute

any of the normal vectors calculated in the previous examples.